\section{Flight Profile Calculations} % (fold)
\label{sec:Flight Profile Calculations}

The balloon characteristics are given in \cite{kaymont}. We now calculate the ascent and descent rates and the burst altitude (should the float system fail).

\subsection{Ascent Rate} % (fold)
\label{ssub:Ascent Rate}

The balloon can be assumed to be in equilibrium during ascent. We can therefore use the equilibrium drag equation (rearranging for velocity):

\[
%
V = \sqrt{\frac{L}{\rho C_d \pi {r_b}^2}}
%	
\]

where $L$ is the `free lift' (the total lift from the helium minus the balloon and payload weight), $\rho$ is the density of the atmosphere, calculated from the NASA Standard Model \cite{nasaatm}, $C_d$ is the coefficient of drag defined by thee Kaymont data, and $r_b$ is the radius of the balloon. We know empirically that the decrease in $\rho$ is almost exactly cancelled by the increase in volume of the balloon (since the compression of the envelope's Helium from elastic forces is so small), so we assume the ascent rate constant. Gravity is assumed constant (it is 1\% weaker at 32km altitude). The balloon is assumed to be a sphere.  Assuming a 5kg payload, and a desired ascent rate of 4ms$^{-1}$, a free lift of \textbf{12.5N} is required. From this we can calculate the burst altitude.

The choice of ascent rate is based upon practical experience - if the ascent rate is too low the balloon tends to drift too far in the fast, lower altitude winds. If the ascent rate is increased then the burst altitude is reduced.

\subsection{Burst Altitude} % (fold)
\label{sub:Burst Altitude}

The balloon ascends until burst, when the volume of gas inside the balloon reaches the burst volume of the balloon. This should be some margin above the desired float altitude to account from deviations from this basic model, such as manufacturing variations between balloons. The Kaymont 3kg balloons have a burst diameter of 13m \cite{kaymont}. For a free lift of 12.5N, a balloon weight of 29.4N and a payload weight of 49N, the sea-level volume of helium required is:

\[
\text{Volume} = \frac{(12.5 + 29.4 + 49)}{(\rho_{air} - \rho_{He})g} = 9.00 \text{m$^3$} 
\]

The balloon's burst volume is 1150 m$^3$, so the balloon will burst at an altitude equivalent to a pressure ratio of 9 $\div$ 1150 = 0.0078, which from \cite{nasaatm} corresponds to an altitude of 35.0km - comfortably above the desired float altitude of 30km.

\subsection{Floating} % (fold)
\label{sub:Floating}

Some experiments have been carried out to confirm that latex balloons can be made to float. A 1.5kg balloon was fitted with a 1.5mm vent and launched such that a desired altitude was reached at sunset. At this point the balloon altitude began to level off and neutral buoyancy was achieved, as can been seen in Figure \ref{fig:floatgraph} (note that unfortunately the original data from that flight has been lost due to a hard drive failure, so a better quality graph cannot be produced, but the effect is demonstrated).   

\begin{figure}[!htbp]

  \centering
  \subfloat[Graph of altitude against time]{\label{fig:floatgraph}\includegraphics[width=0.48\textwidth]{appdx/float.png}}
  \subfloat[Author pictured inflating balloon before fitting experimental floating valve]{\label{mefilling}\includegraphics[width=0.48\textwidth]{appdx/floatfill.jpg}}
  \label{fig:floatingfig}
  \caption{A flight in late summer 2009 confirmed the ability to induce balloon to float at night using a simple valve installed in the balloon neck. Gaps in data are assumed to be because of a GPS failure at low temperature. Lock was resumed at sunrise, at which point the balloon began to ascend to burst.}
\end{figure}

% subsection Floating (end)

% subsection Burst Altitude (end)


% section Flight Profile Calculations (end)
\newpage 
